统计所有小于非负整数 n 的质数的数量。
示例 1:
输入:n = 10
输出:4
解释:小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。
示例 2:
输入:n = 0
输出:0
示例 3:
输入:n = 1
输出:0
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/count-primes
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埃式筛
import java.util.Scanner;
class Solution {
private static int[] cnt = new int[5000001];
static {
boolean[] isNotPrime = new boolean[5000001];
isNotPrime[0] = true;
isNotPrime[1] = true;
for (int i = 4; i <= 5000000; i += 2) {
isNotPrime[i] = true;
}
for (int i = 3; i <= 5000000; i += 2) {
if (!isNotPrime[i]) {
if ((long) i * i <= 5000000) {
for (int j = i * i; j <= 5000000; j += i) {
isNotPrime[j] = true;
}
}
}
}
for (int i = 1; i <= 5000000; ++i) {
if (!isNotPrime[i]) {
cnt[i] = cnt[i - 1] + 1;
} else {
cnt[i] = cnt[i - 1];
}
}
}
public int countPrimes(int n) {
if (n == 0) {
return 0;
}
return cnt[n - 1];
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
System.out.println(new Solution().countPrimes(in.nextInt()));
}
}
}
线性筛
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
class Solution {
private static int[] cnt = new int[5000001];
static {
boolean[] isPrime = new boolean[5000001];
Arrays.fill(isPrime, true);
isPrime[0] = false;
isPrime[1] = false;
List<Integer> primes = new ArrayList<>();
for (int i = 2; i <= 5000000; i++) {
if (isPrime[i]) {
primes.add(i);
}
for (int j = 0; j < primes.size() && primes.get(j) * i <= 5000000; ++j) {
isPrime[i * primes.get(j)] = false;
if (i % primes.get(j) == 0) {
break;
}
}
}
for (int i = 1; i <= 5000000; ++i) {
if (isPrime[i]) {
cnt[i] = cnt[i - 1] + 1;
} else {
cnt[i] = cnt[i - 1];
}
}
}
public int countPrimes(int n) {
if (n == 0) {
return 0;
}
return cnt[n - 1];
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
System.out.println(new Solution().countPrimes(in.nextInt()));
}
}
}